Q. 39 - Calculus

Question

Evaluate each of the double integral in the exercise 37-54 as iterated integrals  

$$\begin{gathered} \int {\int }_R(2-3x^2+y^3)dA \\ Where R=\left\{\right(x,y\left)\right|-3\le x\le 2,3\le y\le 5\} \end{gathered}$$

Step-by-Step Solution

Answer

The value of the integral is 113.33units

Step 1: Given information

We are given an integral

$$\begin{gathered} \int {\int }_R(2-3x^2+y^3)dA \\ Where R=\left\{\right(x,y\left)\right|-3\le x\le 2,3\le y\le 5\} \end{gathered}$$

Step 2: Use iterated integral

We get,

$$\begin{gathered} {\int }_3^5{\int }_{-3}^2(2-3x^2+y^2)dxdy \\ ={\int }_3^5{{[2x-x^3+x{y}^2]}^2}_{-3}dy \\ ={\int }_3^5(10-35+5y^2)dy \\ ={{[-25y+\frac{5}{3}{y}^3]}^5}_3 \\ =113.33units \end{gathered}$$